Page 417 - Advanced engineering mathematics

12.6 Applications of Surface Integrals 397

EXAMPLE 12.22
2 2
2
2
We will find the mass and center of mass of the cone z = x + y for x + y ≤ 4if δ(x, y, z) =
2
2
x + y .
Let D be the disk of radius 2 about the origin. Compute
∂z x ∂z y
= and = .
∂x z ∂y z
The mass is

2
2
m = (x + y )dσ

x y
2 2
2 2
= (x + y ) 1 + + dy dx
z 2 z 2
D
2π 2 √
= r 2 2rdrdθ
0 0
2
√ 1 √
= 2 2π r 4 = 8 2π.
4
0
By symmetry of the surface and of the density function, we expect the center of mass to lie
on the z - axis, so x = y = 0. This can be verified by computation. Finally,
1
2
2
z = √ z(x + y )dσ
8 2π

1 x 2 y 2
2
2
2
2
= √ x + y (x + y ) 1 + + dy dx
8 2π D z 2 z 2
1 2π 2
2
= r(r )rdrdθ
8π 0 0
2
1 1 8

= (2π) r 5 = .
8π 5 5
0
The center of mass is (0,0,8/5).
12.6.3 Flux of a Fluid Across a Surface
Suppose a fluid moves in some region of 3-space with velocity V(x, y, z,t). In studying the flow,
it is often useful to place an imaginary surface in the fluid and analyze the net volume of fluid
flowing across the surface per unit time. This is the flux of the fluid across the surface.
Let n(u,v,t) be the unit normal vector to the surface at time t. If we are thinking of flow out
of the surface from its interior, then choose n to be an outer normal, oriented from a point of the
surface outward away from the interior.
Inatimeinterval t the volume of fluid flowing across a small piece j of approximately
equals the volume of the cylinder with base j and altitude V n t, where V n is the component of
V in the direction of n, evaluated at some point of j . This volume (Figure 12.17) is (V n t)A j ,
where A j is the area of j . Because n = 1, V n = V · n. The volume of fluid across j per unit
time is
(V n t)A j
= V n A j = V · n t.
t

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October 14, 2010 14:53 THM/NEIL Page-397 27410_12_ch12_p367-424